Question

Challenge: Use the diagram.

1. Write expressions for $m\angle 1,m\angle 2,\text{and}m\angle 3$ in terms of $x$ and $y$. Explain your reasoning.
2. Use the expressions you wrote in part (a) to show that the sum of the measures of the angles in $\Delta ABC$ is $180°$.

Short answer

a. The measures of the numbered angles are $m\angle 1=x°$, $m\angle 2=180{}^{\circ }-\left(x{}^{\circ }+y°\right)$, and $m\angle 3=y°$.

b. The sum of all the angles of $\Delta ABC$is$180°$ as shown

$\begin{array}{c}m\angle 1+m\angle 2+m\angle 3=x°+y°+180°-\left(x°+y°\right)\\ =x°+y°+180°-x°-y°\\ =180°\end{array}$

Step-by-step solution

a.Step 1. Given Information.

The numbered angles are $\angle 1,\angle 2,\text{and},\angle 3$, which is the internal angles of the triangle are. And, other two angles $x°$ and $y°$ are provided. The provided figure is,

Step 2. Concept used.

Transversal and parallel lines

The angles internal to the two parallel lines, in alternate corners of the transversal are called alternate interior angles. The measure of alternate interior angles are equal. And, the straight angle measure $180°$.

Step 3. Find the measure of angle 1.

From the given figure, $\angle 1$ and angle $x$ are a pair of alternate interior angles. So, $m\angle 1=x°$ .

Step 4. Find the measure of angle 2.

The sum of the linear paired angles is $180°$, $x{}^{\circ }+m\angle 2+y{}^{\circ }=180°$. So, the measure of angle 2 is,

$\begin{array}{c}x°+m\angle 2+y°={180}^{\circ }\\ m\angle 2=180°-\left(x°+y°\right)\end{array}$

Step 5. Find the measure of angle 3.

From the given figure, $\angle 3$ and angle $y$ are a pair of alternate interior angles. So, $m\angle 3=y°$ .

Step 6. Conclusion.

The measures of the numbered angles are $m\angle 1=x°$, $m\angle 2=180{}^{\circ }-\left(x{}^{\circ }+y°\right)$, and $m\angle 3=y°$.

b.Step 1. Given Information.

The measures of the numbered angles are $m\angle 1=x°$, $m\angle 2=180{}^{\circ }-\left(x{}^{\circ }+y°\right)$, and $m\angle 3=y°$.

Step 2. Calculation.

The sum of numbered angles is,

$\begin{array}{c}m\angle 1+m\angle 2+m\angle 3=x°+y°+180°-\left(x°+y°\right)\\ =x°+y°+180°-x°-y°\\ =180°\end{array}$

Step 3. Conclusion.

Hence, the sum of all the angles of $\Delta ABC$is$180°$ as shown,

$\begin{array}{c}m\angle 1+m\angle 2+m\angle 3=x°+y°+180°-\left(x°+y°\right)\\ =x°+y°+180°-x°-y°\\ =180°\end{array}$